Apiece of metal weighing 59.047 g was heated to 100.0 °c and then put it into 100.0

The specific heat of metal is 0.403J/g°C

Explanation:

To calculate the mass of water, we use the equation:

Density of water = 1 g/mL

Volume of water = 100.0 mL

Putting values in above equation, we get:

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

The equation used to calculate heat released or absorbed follows:

      ......(1)

where,

q = heat absorbed or released

= mass of metal = 59.047 g

= mass of water = 100 g

= final temperature = 27.8°C

= initial temperature of lead = 100°C

= initial temperature of water = 23.7°C

= specific heat of lead = ?

= specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

Hence, the specific heat of metal is 0.403J/g°C

Mg (s) + 2 HCl (aq) -> MgCl2 (aq) + H2 (g)

1) Number of mols of HCl, n

n = 6.3 L * 4.5 mol/L = 28.35 mol

2) ratio: 2 mol HCl / 1 mol Mg = 28.35 mol HCl / x mol Mg

x = 28.35 mol HCL * 1 mol Mg / 2 mol HCL = 14.175 mol Mg

3) Convert mol to mass using atomic mass of Mg

14.175 mol Mg * 24.3 g Mg / mol Mg = 344. 45 g

344.45 g